Question

Se han muestreado vapores de alcohol isopropílico usando un tubo de carbón activado durante 8 horas a un flujo de 20 ml/min: ¿cuál fue el volumen del aire muestreado?

Answer 1

For this case we can translate the given statement as:

Isopropyl alcohol vapors have been sampled using an activated carbon tube for 8 hours at a flow of 20 ml/min: what was the volume of the sampled air?

By definition, the volumetric flow rate is given by:

`Q = \frac {V} {t}`

Where:

V: It's the volume

t: It's time

Then, the volume is given by:
`V = Q * t`

According to the problem data we have:

`Q = 20 \frac {ml} {min}\\t = 8h * \frac {60} {1} \frac {min} {h} = 480min`

Substituting the values we have:

`V = 20 \frac {ml} {min} * 480min\\V = 9600ml`

Finally, the volume of the sample is:

9600ml

Answer:

9600ml